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Sergio [31]
4 years ago
15

Number one and number two

Mathematics
1 answer:
Amanda [17]4 years ago
4 0
1.  sum of exterior angles = 360 

so one interior angles = 180 - (360/30) = 180 - 12 = 168 degrees
sum of all the interior angles = 168*30 =  5040  
B


2.   H  

the sum of exterior angles of all polygons is 360 degrres

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Identify the base of the parallelogram, given that A=81x2 cm2.
melamori03 [73]
Ans: Third option, b= 18x cm

Base = Area / Height = 81x^2 / 4.5x
= 18x cm

Finding a parallelogram’s area is essentially finding a rectangle’s area. See below:

4 0
3 years ago
Read 2 more answers
ABCD is a rectangle in which O is the point of intersection of the diagonals. Also, O is equidistant from A, B, C. Is this true?
slavikrds [6]

Answer:

The correct answer is yes its true.

Step-by-step explanation:

Let ABCD is a rectangle which means rectangle have opposite sides of same length : AB = CD and AD = BC.

It is given that O is the point of intersection of the diagonals that means O is mid point of AC and DB,

Then, OA = OB = OC = OD which means O is at similiar distance from each A, B, C and D.

Thus, the correct answer is : True.

8 0
3 years ago
NEED HELP ASAP PLEASE IDK AND IT'S TIMED pls help meh​
aivan3 [116]

Answer: 29/5

Step-by-step explanation:

[3(x-1)]/24 = (x-4)/3

(x-1)/8 = (x-4)/3

3(x-1) = 8(x-4)

3x-3 = 8x -32

3x - 8x = -32 +3

-5x = -29

x = 29/5

7 0
3 years ago
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A-1 Pizza sells 2-liter bottles of soda that they have marked up 150 percent above their wholesale cost of $0.80 per bottle. The
aleksley [76]
.80 x 150% +9% = 2.18
3 0
4 years ago
Read 2 more answers
Find general solutions of the differential equation. Primes denote derivatives with respect to x.
zepelin [54]

Answer:

\mathbf{3x^2y^3+2xy^4=C}

Step-by-step explanation:

From the differential equation given:

6xy^3 +2y ^4 +(9x^2y^2+8xy^3) y' = 0

The equation above can be re-written as:

6xy^3 +2y^4 +(9x^2y^2+8xy^3)\dfrac{dy}{dx}=0

(6xy^3 +2y^4)dx +(9x^2y^2+8xy^3)dy=0

Let assume that if function M(x,y) and N(x,y) are continuous and have continuous first-order partial derivatives.

Then;

M(x,y) dx + N (x,y)dy = 0; this is exact in R if and only if:

\dfrac{{\partial M }}{{\partial y }}= \dfrac{\partial N}{\partial  x}}} \ \ \text{at each point of R}

relating with equation M(x,y)dx + N(x,y) dy = 0

Then;

M(x,y) = 6xy^3 +2y^4\  and \ N(x,y) = 9x^2 y^2 +8xy^3

So;

\dfrac{\partial M}{\partial y }= 18xy^3 +8y^3

       \dfrac{\partial N}{\partial y }

Let's Integrate \dfrac{\partial F}{\partial x}= M(x,y) with respect to x

Then;

F(x,y) = \int (6xy^3 +2y^4) \ dx

F(x,y) = 3x^2 y^3 +2xy^4 +g(y)

Now, we will have to differentiate the above equation with respect to y and set \dfrac{\partial F}{\partial x}= N(x,y); we have:

\dfrac{\partial F}{\partial y} = \dfrac{\partial}{\partial y } (3x^2y^3+2xy^4+g(y)) \\ \\ = 9x^2y^2 +8xy^3 +g'(y) \\ \\ 9x^2y^2 +8xy^3 +g'(y) =9x^2y^2 +8xy^3 \\ \\ g'(y) = 0  \\ \\ g(y) = C_1

Hence, F(x,y) = 3x^2y^3 +2xy^4 +g(y)  \\ \\ F(x,y) = 3x^2y^3 + 2xy^4 +C_1

Finally; the general solution to the equation is:

\mathbf{3x^2y^3+2xy^4=C}

5 0
3 years ago
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